Belarus 1999

[SANZEED]

Let $a,b,c$ be positive real numbers such that $a^{2}+b^{2}+c^{2}=3$. Prove that,
$\frac{1}{1+ab}+\frac{1}{1+bc}+\frac{1}{1+ca}\geq \frac{3}{2}$

[SANZEED]

Let us assume that $f(x)=\frac{1}{1+x}$. Then $f"(x)=\frac{2}{(1+x)^{3}}>0$ for positive $x$. So the function is convex for positive $x$. So, $f(ab)+f(bc)+f(ca)\geq 3f(\frac{ab+bc+ca}{3})=3\frac{1}{1+\frac{ab+bc+ca}{3}}=\frac{3(a^{2}+b^{2}+c^{2})}{a^{2}+b^{2}+c^{2}+ab+bc+ca}$. It suffices to show that $\frac{a^{2}+b^{2}+c^{2}}{a^{2}+b^{2}+c^{2}+ab+bc+ca}\geq \frac{1}{2}$, i.e. $a^{2}+b^{2}+c^{2}\geq ab+bc+ca$ which follows from AM-GM. Equality holds for $a=b=c$.